Problem: Multiply the following complex numbers: $({-4-2i}) \cdot ({1-3i})$
Answer: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({-4-2i}) \cdot ({1-3i}) = $ $ ({-4} \cdot {1}) + ({-4} \cdot {-3}i) + ({-2}i \cdot {1}) + ({-2}i \cdot {-3}i) $ Then simplify the terms: $ (-4) + (12i) + (-2i) + (6 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -4 + (12 - 2)i + 6i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -4 + (12 - 2)i - 6 $ The result is simplified: $ (-4 - 6) + (10i) = -10+10i $